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With this information, the former constraint is trivial.
Aξ x ψ x + B(ξ x ψ y + ξ y ψ x ) + Cξ y ψ y = 0
Aξ x (−B/A) + B(ξ x + ξ y (−B/A)) + Cξ y = 0
2
(AC − B )ξ y = 0
0 = 0

Thus we have a first order partial differential equation for the ψ coordinate which we can
solve with the method of characteristics.
B
ψ x + ψ y = 0
A
The ξ coordinate is chosen to be anything linearly independent of ψ. The characteristic
equations for ψ are
dy B d
= , ψ(x, y(x)) = 0
dx A dx
Solving the differential equation for y(x) determines ψ(x, y). We just write the solution
for y(x) in the form F(x, y(x)) = const. Since the solution of the differential equation
for ψ is ψ(x, y(x)) = const, we then have ψ = F(x, y). Upon solving for ψ and choosing
∗
a linearly independent ξ, we divide Equation 3.4 by A to obtain the canonical form.
In the case that A = 0, we would instead have the constraint,
B
ψ x + ψ y = 0.
C
Example 3.4 Prove that the equation

2
2
x u xx − 2xyu xy + y u yy + xu x + yu y = 0 (3.15)
is parabolic and find its canonical form; find the general solution on the half-plane
x > 0.
2
2
2 2
2 2
2
We identify a = x , 2b = −2xy, c = y ; therefore, b − ac = x y − x y = 0 and
y
the equation is parabolic. The equation for the characteristics is dy = − , and the
dx x
solution is xy = constant. Therefore, we define ψ(x, y) = xy. The second variable
can be simply chosen as ξ(x, y) = x. Let ν(ξ, ψ) = u(x, y). Substituting the new
coordinates ξ and ψ into (3.15), we obtain
2
2
2
x (y ν ψψ + 2yν ξψ + ν ξξ ) − 2xy(ν ψ + xyν psiψ + xν ξψ ) + x ν ψψ + xyν ψ + xν ξ + xyν ξ = 0.
Thus,
2
ξ ν ξξ + ξν ξ = 0,
or ν ξξ + (1/ξ)ν ξ = 0, and this is the desired canonical form.
Setting w = ν ξ , we arrive at the first-order ODE w ξ + (1/ξ)w = 0. The solution
is ln w = − ln ξ + f(ψ), or w = f(ψ)/ξ. Hence, v satisfies
Z Z Z
f(ψ)
ν = ν ξ dξ = wdξ = dξ = f(ψ) ln ξ + g(ψ).
ξ
Therefore, the general solution u(x, y) of (3.15) is u(x, y) = f(xy) ln x+g(xy), where
f, g are arbitrary real twice continuously differentiable functions.

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